1 VukazichCE160Lab2Notes[L2]

CE 160 Lab 2 Notes: Shear and Moment Diagrams for Beams

Shear and moment diagrams are plots of how the internal bending moment and shear vary along the length of the beam.

Sign Convention for V and M

Consider the arbitrarily loaded simple beam:

Cut the beam loose from its pin support at the left, roller support at the right and through the beam at section a-a

Free Body Diagrams (F. B. D.) of portion of beam to the left of a-a and portion of the beam to the right of a-a (note that there is no axial force in this beam):

Reactive forces are developed at each support and equal and opposite internal shear force (V) and bending moment (M) are present to keep each beam segment in equilibrium.

The above senses for V and M are the “usual” Civil Engineering convention for positive shear and positive bending moment.

It is good practice to show the sign convention next to shear and moment diagrams to make the user clear of the convention for positive shear and moment.

a

a

a

aM

a

aM

V V

2 VukazichCE160Lab2Notes[L2]

Some common icons for showing sign convention

Example Problem

Construct the Shear and Bending Moment Diagram for the following beam:

Find support reactions at A and C

F.B.D. of entire beam cut loose from its supports

Need to replace distributed load as an equivalent point load (equivalent point load = area of distribution placed at the centroid of the distribution)

A

B

C

10 ft 10 ft

10 k2 k/ft

Ax

B

C

10 ft 10 ft

10 k2 k/ft

Ay Cy

A

3 VukazichCE160Lab2Notes[L2]

Apply equations of equilibrium to find unknown reactions

Moment Equilibrium

𝑀! = 0

Counterclockwise moments about point A positive

− 20 𝑘×5 𝑓𝑡 + − 10 𝑘 ×10 𝑓𝑡 + 𝐶! ×20 𝑓𝑡 = 0 𝑪𝒚 = 𝟏𝟎 𝒌 (Cy is positive so Cy acts upward as assumed)

Force Equilibrium

+↑ 𝐹! = 0

Upward forces positive

𝐴! − 20𝑘 − 10 𝑘 + 𝐶! = 0

𝑨𝒚 = 𝟐𝟎 𝒌 (Ay is positive so Ay acts upward as assumed)

Force Equilibrium

+→ 𝐹! = 0

forces positive to the right

𝐴! = 0

𝑨𝒙 = 𝟎 (no axial force in beam)

Ax

B

C

5 ft 10 ft

10 k

(2 k/ft)*(10 ft) = 20 k

Ay Cy

A

5 ft

+

4 VukazichCE160Lab2Notes[L2]

F.B.D of beam with known support reactions shown

Find internal shear and moment from A to B

Make a cut at section a-a, an arbitrary point between points A and B F.B.D of beam to left of cut, show unknown internal forces V and M in their positive sense

Replace distributed load as an equivalent point load

B

C

10 ft 10 ft

10 k2 k/ft

20 k 10 k

A

a

a

b

b

2 k/ft

A

20 k x

M

Va

a

A

20 k x/2

M

Va

a(2 k/ft)*(x) = 2x

x/2

5 VukazichCE160Lab2Notes[L2]

Moment Equilibrium of segment

𝑀! = 0

Counterclockwise moments about point a are positive

− 20×𝑥 + 2𝑥×𝑥/2 + 𝑀 = 0 𝑴 = 𝟐𝟎𝒙− 𝒙𝟐 (units in k and ft)

Force Equilibrium

+↑ 𝐹! = 0

Upward forces positive

20 − 2𝑥 − 𝑉 = 0

𝑽 = 𝟐𝟎− 𝟐𝒙 (units in k and ft)

Plot V and M between points A and B

At point A Evaluate at x = 0

𝑉! = 20− 2×0 = 𝟐𝟎 𝒌 𝑀! = 20×0 − 0! = 𝟎 𝒌− 𝒇𝒕

At point B- (just to the left of point B) Evaluate at x = 10ft

𝑉! = 20− 2×10 = 𝟎 𝒌 𝑀! = 20×10 − 10! = 𝟏𝟎𝟎 𝒌− 𝒇𝒕

V is linear in x between A and B

M is parabolic in x between A and B

+

6 VukazichCE160Lab2Notes[L2]

Find internal shear and moment from B to C Make a cut at section b-b, an arbitrary point between points B and C

F.B.D of beam to right of cut, show unknown internal forces V and M in their positive sense

Moment Equilibrium of segment

𝑀! = 0

Counterclockwise moments about point b are positive

10 ×𝑥′ − 𝑀 = 0 𝑴 = 𝟏𝟎𝒙! (units in k and ft)

Force Equilibrium

+↑ 𝐹! = 0

Upward forces positive

10 + 𝑉 = 0 𝑽 = −𝟏𝟎 (units in k)

Plot V and M between points B and C

At point B+ (just to the right of point B) Evaluate at x’ = 10ft

𝑉! = −𝟏𝟎 𝒌 𝑀! = 10×10 = 𝟏𝟎𝟎 𝒌− 𝒇𝒕

x’

C

10 kb

b

M

V

+

7 VukazichCE160Lab2Notes[L2]

At point C Evaluate at x’ = 0

𝑉! = −𝟏𝟎 𝒌 𝑀! = 10×0 = 𝟎 𝒌− 𝒇𝒕

V is constant between B and C

M is linear in x’ between B and C

Plot V and M diagrams under the F.B.D of the beam

M

B

C

10 ft 10 ft

10 k2 k/ft

20 k 10 k

A

x x’

V-10 k

20 k

0

100 k-ft

0

220 xxM −=

'10xM =

xV 220 −=

10−=V

8 VukazichCE160Lab2Notes[L2]

For review, we will use equilibrium to find the variation of V and M across the beam in this lab. However, we should note the differential and integral relationships between V, M, and w that are also useful. Note also that these relationships are dependent on the sign convention chosen.

𝑑𝑉𝑑𝑥 = 𝑤 slope of tangent to V diagram at a point

= (distributed load intensity at that point)

𝑑𝑀𝑑𝑥 = 𝑉 slope of tangent to M diagram at a point = value of V at that point

VB – VA = - (Area of distributed load between points A and B)

MB – MA = Area of V diagram between points A and B